SOLUTION: 5(x+3)+9 < or = to 3 (x-2)+6 5x+15+9 < or = to 3x-6+6 5x+24 < or = to 3x 5x/3x+24 < or = to 3x/3x this is where I get hung up what do I do with the 5x/3x?

Algebra ->  Algebra  -> Equations -> SOLUTION: 5(x+3)+9 < or = to 3 (x-2)+6 5x+15+9 < or = to 3x-6+6 5x+24 < or = to 3x 5x/3x+24 < or = to 3x/3x this is where I get hung up what do I do with the 5x/3x?       Log On

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Question 17960: 5(x+3)+9 < or = to 3 (x-2)+6
5x+15+9 < or = to 3x-6+6
5x+24 < or = to 3x
5x/3x+24 < or = to 3x/3x
this is where I get hung up what do I do with the 5x/3x?

Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
Well, you got started ok but your fourth step is in error. You must collect the x-terms (5x & 3x) together on the same side of the = sign.
5x+%2B+24+%3C=+3x At this point, you can subtract 24 from both sides.
5x+%3C=+3x+-+24 Now, subtract 3x from both sides.
2x+%3C=+-24 Finally, divide both sides by 2.
x+%3C=+-12